Projectile Problem Homework: Solve Range, Time, Velocity, Height

  • Thread starter munkachunka
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In summary, the mortar shell fired with an angle of elevation of 52 deg has a range of 2.1km, neglecting air resistance.
  • #1
munkachunka
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Homework Statement



I have made an attempt at the following could someone please check my work.

a mortar shell fired with an angle of elevation of 52 deg has a range of 2.1km, neglecting air resistance, determine.

1, time of flight
2, initial velocity
3, shells range and height after 10 secs
4, the other range at which the height was the same as calculated in part 3 above

Homework Equations



x=u(cos(angle))t
h= u(sin, (angle))t-1/2gt^2

The Attempt at a Solution



a, rearanging x=u(cos52)t to equal x/cos52=ut
1hen x = 2.5km h=0

h=u(sin angle)t-1/2gt^2

substituting 0=(2.5knm/cos52)*(sin52)-1/2.9.8.t^2

transposing for t = 23.409

2, x=u(cos52)t therefore u=2100/(cos52)*23.409 = 145.714m/s

3= x=u(cos52)t = 145.712(cos52).10 = 897.093M

height = u(sin52)t-1/2gt^2
= 145.712*(sin52)*10-1/2*9.8*10^2

height = 658.226

4 assuming angle of firing = angle of descent and initial acceleration = rate of deceleration then the other range = 2.1km-897.093 = 1202.907

assuming the first 3 parts are correct then the part I am not too sure about is part 4, there are 12 marks allocated to this and so looks too easy.

your time is greatly appreciated
 
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  • #2
munkachunka said:
assuming the first 3 parts are correct then the part I am not too sure about is part 4, there are 12 marks allocated to this and so looks too easy.
It looks good to me. (Just round off your answers to a reasonable number of significant figures.)

If you are unsure about part 4, solve it a second way and compare. (Use your equation for height as a function of time, solve for the time, plug into your equation for x to find the range.)
 
  • #3
Yeah, everything looks good... Just noticed this part:

0=(2.5knm/cos52)*(sin52)-1/2.9.8.t^2

the problem says 2.1km, you left a t out (probably nothing to do with your written work, but just typos)... but your calculations are right...

I noticed I was getting a few decimal places off from you in the answers... but other than that everything was the same.
 
Last edited:
  • #4
thankyou for your time
 

1. What is a projectile?

A projectile is an object that is launched into the air and moves along a curved path due to the force of gravity acting upon it.

2. How do you calculate the range of a projectile?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the formula R = v0cosθ * t, where R is the range, v0 is the initial velocity, θ is the launch angle, and t is the time the projectile is in the air.

3. What is the relationship between the launch angle and the range of a projectile?

The range of a projectile is directly proportional to the sine of twice the launch angle. This means that the larger the launch angle, the greater the range of the projectile. However, if the launch angle is too large (above 90 degrees), the projectile will not travel horizontally and will instead fall straight down.

4. How do you calculate the time of flight for a projectile?

The time of flight for a projectile is the amount of time it takes to reach its peak height and then return to the ground. It can be calculated using the formula t = 2v0sinθ/g, where v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (9.8 m/s2).

5. What is the relationship between the initial velocity and the maximum height of a projectile?

The maximum height of a projectile is directly proportional to the square of the initial velocity. This means that the higher the initial velocity, the greater the maximum height the projectile will reach. However, the angle of launch also affects the maximum height, with a 45 degree launch angle resulting in the greatest height for a given initial velocity.

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